Question

Question #a83c1

Answer 1

The reaction was initiated by volimetric ratio of Hydrigen and Nitrogen as `3:1` and at equilibrium the ratio remained more or less same #36%:13%~~3:1 # (Within experimental error.)
So the volume of `NH_3` in the equilibrium mixture may be taken as `(100-36-13)%=51%`

If the pressure of the reaction mixture at equilibrium be P atm then the partial pressures of the constituents of the reaction mixture will be as follows.

Partial presssure of Hydrogen
`p_(H_2)=36%ofP=0.36P" atm"`

Partial presssure of Nitrogen
`p_(N_2)=13%ofP=0.13P" atm"`

Partial presssure of Ammonia
`p_(NH_3)=51%ofP=0.51P" atm"`

The equation of the gaseous reaction:

`3H_2(g)+N_2(g)rightleftharpoons2NH_3(g)`

Now the equlibrium constant of the gaseous reaction in respect of pressure is given by

`K_p=(p_(NH_3))^2/(p_(N_2)xx(p_(H_2))^3)`

`=>K_p=(0.51P)^2/(0.13Pxx(0.36P)^3)" atm"^-2`

`~~42.88P^-2" atm"^-2`

If the value of P is known then `K_p` can be calculated out.